Rotational Motion of a shaft

  • Thread starter Chuck 86
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  • #1
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Homework Statement


A shaft is turning at 73.8 rad/s at time zero. Thereafter, its angular acceleration is given by
a = -10.1 rad/s2 - 4.44 t rad/s3.
where t is the elapsed time. Calculate its angular speed at t = 2.58 s

Homework Equations


1/2(Angular accel formula)+ 73.8t+0= theta final


The Attempt at a Solution


do i take the integral of the accel formula then the velocity the plug in 2.58?
 

Answers and Replies

  • #2
kuruman
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You need to integrate both sides of

dω = αdt

The limits of integration should be

73.8 to ω (left side) and 0 to 2.58 (right side)

After you integrate, solve for ω.
 
  • #3
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so i set 1/2(angular accel)= 73.8+ theta final? then take the integral?
 
  • #4
kuruman
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so i set 1/2(angular accel)= 73.8+ theta final? then take the integral?
I never said that. I have no idea where you get that 1/2(angular accel) = 73.8 + theta final means something when it is not even dimensionally correct. Do the following integral and solve for ω. That's what I suggested in the first place.

[tex]\int_{73.8}^{\omega}d \omega=\int_{0}^{2.58}(-10.1-4.44 t) dt[/tex]
 
  • #5
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whoa dude calm down. i appericiate the help but remember i was just asking a question, i already feel stupid enough as it is
 

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